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In this java program explain every line of this code. Thank you Source Code: import java.util.Scanner; public class Main {  public static void main(String[] args) {      Scanner in = new Scanner(System.in);   System.out.println("The function is f(x)= 2x+3(2)");   System.out.println("Width is 0.05");   System.out.print("Enter Lower Limit: ");   double lowLim = in.nextFloat();   System.out.print("Enter Upper Limit: ");   double upperLim = in.nextFloat();   double iterations = (upperLim - lowLim) / 0.05;

In this java program explain every line of this code. Thank you Source Code: import java.util.Scanner; public class Main {  public static void main(String[] args) {      Scanner in = new Scanner(System.in);   System.out.println("The function is f(x)= 2x+3(2)");   System.out.println("Width is 0.05");   System.out.print("Enter Lower Limit: ");   double lowLim = in.nextFloat();   System.out.print("Enter Upper Limit: ");   double upperLim = in.nextFloat();   double iterations = (upperLim - lowLim) / 0.05;

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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In this java program explain every line of this code. Thank you

Source Code:

import java.util.Scanner;

public class Main
{
 public static void main(String[] args) {
     Scanner in = new Scanner(System.in);
  System.out.println("The function is f(x)= 2x+3(2)");
  System.out.println("Width is 0.05");
  System.out.print("Enter Lower Limit: ");
  double lowLim = in.nextFloat();
  System.out.print("Enter Upper Limit: ");
  double upperLim = in.nextFloat();
  double iterations = (upperLim - lowLim) / 0.05;
  int i = 0;
  double finAns = 0;
  while(i < iterations){
      System.out.print("____________________________________________________________________________________________________________________________________");
      System.out.print("\nIteration:" + (i + 1));
      System.out.print("\nFunction\t|\tLower Limit \t|\t Upper Limit \t|\t LRAM Application (width=0.05) \t|\t Answer");
      double initAns=((2*lowLim)+(3*2))*0.05;
      System.out.printf("\nf(x)= 2x+3(2) \t\t|   x=%.2f\t|\tx=%.2f\t   |", lowLim,lowLim+0.05);
      System.out.printf("\t f(x) = (2(%.2f)+3(2) * 0.05\t    |\t       %.2f\n",lowLim,initAns);
      finAns += initAns;
      lowLim+=0.05;
      i += 1;
  }
  System.out.println("____________________________________________________________________________________________________________________________________");
  System.out.println("Final answer/Summation of all iterations is:" + finAns);
 }
}

 

..

See the picture below that's a output of this program.

Thank you

The function is f(x)= 2x+3(2) Width is 0.05 Enter Lower Limit: 1 Enter Upper Limit: 2 Iteration: 1 Function f(x)= 2x+3(2) ! Upper Limit x=1.05 Lower Limit LRAM Application (width=0.05) f(x) = (2(1.00)+3(2) Answer 0.40 x=1.00 * 0.05 Iteration: 2 Lower Limit x=1.05 | LRAM Application (width=0.e5) | f(x) = (2(1.e5)+3(2) • 0.05 Function Upper Limit Answer f(x)= 2x+3(2) X=1.10 0.41 Iteration: 3 LRAM Application (width=0.05) f(x) = (2(1.10)+3(2) * 0.05 Function Lower Limit Upper Limit Answer f(x)= 2x+3(2) X=1.10 x=1.15 0.41 Iteration: 4 Lower Limit I x=1.15 LRAM Application (width=0.05) | f(x) = (2(1.15)+3(2) * 0.05 Function Upper Limit Answer f(x)= 2x+3(2) x=1.20 0.42 Iteration: 5 Upper Limit LRAM Application (width=0.05) | f(x) = (2(1.20)+3(2) * 0.05 Function Lower Limit Answer f(x)= 2x+3(2) x=1.20 x=1.25 0.42 Iteration: 6 Function f(x)= 2x+3(2) LRAM Application (width=0.05) I f(x) = (2(1.25)+3(2) * 0.05 Lower Limit Upper Limit x=1.30 Answer X=1.25 0.43 Iteration: 7 Function f(x)= 2x+3(2) Lower Limit x=1.30 Upper Limit x=1.35 LRAM Application (width=0.05) | F(x) = (2(1.30)+3(2) * 0.05 Answer 0.43 Iteration: 8 Function f(x)= 2x+3(2) Upper Limit I LRAM Application (width=e.05) | f(x) = (2(1.35)+3(2) * 0.05 Lower Limit ! x=1.40 Answer x=1.35 0.44 Iteration: 9 Function f(x)= 2x+3(2) Upper Limit x=1.45 LRAM Application (width=0.05) | f(x) = (2(1.40)+3(2) * 0.05 Lower Limit Answer x=1.40 0.44 Iteration: 10 Function f(x)= 2x+3(2) Upper Limit x-1.50 LRAM Application (width=0.05) | f(x) = (2(1.45)+3(2) * 0.05 Lower Limit Answer x=1.45 0.45 Iteration: 11 Lower Limit | x=1.50 ! Upper Limit x=1.55 Function | LRAM Application (width=0.05) | Answer f(x)= 2x+3(2) f(x) = (2(1.50)+3(2) * 0.05 I 0.45 Iteration: 12 Function f(x)= 2x+3(2) LRAM Application (width=0.05) | f(x) = (2(1.55)+3(2) * 0.05 Lower Limit Upper Limit Answer x=1.55 X=1.60 ө.46 Iteration: 13 Function f(x)= 2x+3(2) LRAM Application (width=0.05) | f(x) = (2(1.60)+3(2) * 0.05 Lower Limit ! Upper Limit Answer x=1.60 x=1.65 0.46 ----. 14
Transcribed Image Text:The function is f(x)= 2x+3(2) Width is 0.05 Enter Lower Limit: 1 Enter Upper Limit: 2 Iteration: 1 Function f(x)= 2x+3(2) ! Upper Limit x=1.05 Lower Limit LRAM Application (width=0.05) f(x) = (2(1.00)+3(2) Answer 0.40 x=1.00 * 0.05 Iteration: 2 Lower Limit x=1.05 | LRAM Application (width=0.e5) | f(x) = (2(1.e5)+3(2) • 0.05 Function Upper Limit Answer f(x)= 2x+3(2) X=1.10 0.41 Iteration: 3 LRAM Application (width=0.05) f(x) = (2(1.10)+3(2) * 0.05 Function Lower Limit Upper Limit Answer f(x)= 2x+3(2) X=1.10 x=1.15 0.41 Iteration: 4 Lower Limit I x=1.15 LRAM Application (width=0.05) | f(x) = (2(1.15)+3(2) * 0.05 Function Upper Limit Answer f(x)= 2x+3(2) x=1.20 0.42 Iteration: 5 Upper Limit LRAM Application (width=0.05) | f(x) = (2(1.20)+3(2) * 0.05 Function Lower Limit Answer f(x)= 2x+3(2) x=1.20 x=1.25 0.42 Iteration: 6 Function f(x)= 2x+3(2) LRAM Application (width=0.05) I f(x) = (2(1.25)+3(2) * 0.05 Lower Limit Upper Limit x=1.30 Answer X=1.25 0.43 Iteration: 7 Function f(x)= 2x+3(2) Lower Limit x=1.30 Upper Limit x=1.35 LRAM Application (width=0.05) | F(x) = (2(1.30)+3(2) * 0.05 Answer 0.43 Iteration: 8 Function f(x)= 2x+3(2) Upper Limit I LRAM Application (width=e.05) | f(x) = (2(1.35)+3(2) * 0.05 Lower Limit ! x=1.40 Answer x=1.35 0.44 Iteration: 9 Function f(x)= 2x+3(2) Upper Limit x=1.45 LRAM Application (width=0.05) | f(x) = (2(1.40)+3(2) * 0.05 Lower Limit Answer x=1.40 0.44 Iteration: 10 Function f(x)= 2x+3(2) Upper Limit x-1.50 LRAM Application (width=0.05) | f(x) = (2(1.45)+3(2) * 0.05 Lower Limit Answer x=1.45 0.45 Iteration: 11 Lower Limit | x=1.50 ! Upper Limit x=1.55 Function | LRAM Application (width=0.05) | Answer f(x)= 2x+3(2) f(x) = (2(1.50)+3(2) * 0.05 I 0.45 Iteration: 12 Function f(x)= 2x+3(2) LRAM Application (width=0.05) | f(x) = (2(1.55)+3(2) * 0.05 Lower Limit Upper Limit Answer x=1.55 X=1.60 ө.46 Iteration: 13 Function f(x)= 2x+3(2) LRAM Application (width=0.05) | f(x) = (2(1.60)+3(2) * 0.05 Lower Limit ! Upper Limit Answer x=1.60 x=1.65 0.46 ----. 14
Upper Limit x=1.35 |Answer Function T T LRAM Application (width=0.05) f(x) = (2(1.30)+3(2) * 0.05 Lower Limit f(x)= 2x+3(2) X=1.30 0.43 Iteration: 8 ! Upper Limit x=1.40 Function Lower Limit | x=1.35 LRAM Application (width=0.05) | f(x) = (2(1.35)+3(2) * 0.05 Answer f(x)= 2x+3(2) 0.44 Iteration: 9 ! Upper Limit x=1.45 | LRAM Application (width=0.05) | Function Lower Limit Answer 0.44 f(x)= 2x+3(2) x=1.40 f(x) = (2(1.40)+3(2) ө.05 Iteration: 10 Upper Limit x-1.50 LRAM Application (width=0.e5) I f(x) = (2(1.45)+3(2) * 0.05 Function Lower Limit Answer f(x)- 2х+3(2) X=1.45 0.45 Iteration: 11 I Lower Limit x=1.50 | Upper Limit x=1.55 LRAM Application (width=0.05) | f(x) = (2(1.50)+3(2) * 0.05 Function Answer f(x)= 2x+3(2) 0.45 Iteration: 12 I LRAM Application (width=e.05) | f(x) = (2(1.55)+3(2) Function Lower Limit ! Upper Limit Answer 0.46 f(x)= 2x+3(2) x=1.55 x=1.60 0.05 Iteration: 13 Lower Limit | x=1.60 Upper Limit x=1.65 LRAM Application (width=0.e5) | f(x) = (2(1.60)+3(2) * 0.05 Function Answer f(x)= 2x+3(2) ө.46 Iteration: 14 Upper Limit x=1.70 | LRAM Application (width=0.e5) | f(x) = (2(1.65)+3(2) * 0.05 Function Lower Limit Answer f(x)= 2x+3(2) x=1.65 0.47 Iteration: 15 Function LRAM Application (width=0.05) I f(x) = (2(1.70)+3(2) * 0.05 Lower Limit Upper Limit Answer f(x)= 2x+3(2) x=1.70 x=1.75 0.47 Iteration: 16 Function f(x)= 2x+3(2) Lower Limit Upper Limit x=1.80 LRAM Application (width=0.e5) I F(x) = (2(1.75)+3(2) * 0.05 Answer x=1.75 0.48 Iteration: 17 Function Upper Limit x-1.85 | LRAM Application (width=0.e5) | f(x) = (2(1.8e)+3(2) • 0.05 Lower Limit Answer f(x)= 2x+3(2) X=1.80 0.48 Iteration: 18 Function Lower Limit Upper Limit LRAM Application (width=0.05) | Answer 0.49 f(x)= 2x+3(2) x=1.85 x=1.90 F(x) = (2(1.85)+3(2) * 0.05 Iteration: 19 Function f(x)= 2x+3(2) ! Upper Limit X=1.95 LRAM Application (width=0.05) | f(x) = (2(1.90)+3(2) Lower Limit Answer x=1.90 0.49 Iteration: 20 LRAM Application (width=0.05) | f(x) = (2(1.95)+3(2) * 0.05 Function Lower Limit Upper Limit Answer f(x)= 2x+3(2) | x=1.95 x=2.00 0.50 Final answer/Summation of all iterations is: 8.95 --
Transcribed Image Text:Upper Limit x=1.35 |Answer Function T T LRAM Application (width=0.05) f(x) = (2(1.30)+3(2) * 0.05 Lower Limit f(x)= 2x+3(2) X=1.30 0.43 Iteration: 8 ! Upper Limit x=1.40 Function Lower Limit | x=1.35 LRAM Application (width=0.05) | f(x) = (2(1.35)+3(2) * 0.05 Answer f(x)= 2x+3(2) 0.44 Iteration: 9 ! Upper Limit x=1.45 | LRAM Application (width=0.05) | Function Lower Limit Answer 0.44 f(x)= 2x+3(2) x=1.40 f(x) = (2(1.40)+3(2) ө.05 Iteration: 10 Upper Limit x-1.50 LRAM Application (width=0.e5) I f(x) = (2(1.45)+3(2) * 0.05 Function Lower Limit Answer f(x)- 2х+3(2) X=1.45 0.45 Iteration: 11 I Lower Limit x=1.50 | Upper Limit x=1.55 LRAM Application (width=0.05) | f(x) = (2(1.50)+3(2) * 0.05 Function Answer f(x)= 2x+3(2) 0.45 Iteration: 12 I LRAM Application (width=e.05) | f(x) = (2(1.55)+3(2) Function Lower Limit ! Upper Limit Answer 0.46 f(x)= 2x+3(2) x=1.55 x=1.60 0.05 Iteration: 13 Lower Limit | x=1.60 Upper Limit x=1.65 LRAM Application (width=0.e5) | f(x) = (2(1.60)+3(2) * 0.05 Function Answer f(x)= 2x+3(2) ө.46 Iteration: 14 Upper Limit x=1.70 | LRAM Application (width=0.e5) | f(x) = (2(1.65)+3(2) * 0.05 Function Lower Limit Answer f(x)= 2x+3(2) x=1.65 0.47 Iteration: 15 Function LRAM Application (width=0.05) I f(x) = (2(1.70)+3(2) * 0.05 Lower Limit Upper Limit Answer f(x)= 2x+3(2) x=1.70 x=1.75 0.47 Iteration: 16 Function f(x)= 2x+3(2) Lower Limit Upper Limit x=1.80 LRAM Application (width=0.e5) I F(x) = (2(1.75)+3(2) * 0.05 Answer x=1.75 0.48 Iteration: 17 Function Upper Limit x-1.85 | LRAM Application (width=0.e5) | f(x) = (2(1.8e)+3(2) • 0.05 Lower Limit Answer f(x)= 2x+3(2) X=1.80 0.48 Iteration: 18 Function Lower Limit Upper Limit LRAM Application (width=0.05) | Answer 0.49 f(x)= 2x+3(2) x=1.85 x=1.90 F(x) = (2(1.85)+3(2) * 0.05 Iteration: 19 Function f(x)= 2x+3(2) ! Upper Limit X=1.95 LRAM Application (width=0.05) | f(x) = (2(1.90)+3(2) Lower Limit Answer x=1.90 0.49 Iteration: 20 LRAM Application (width=0.05) | f(x) = (2(1.95)+3(2) * 0.05 Function Lower Limit Upper Limit Answer f(x)= 2x+3(2) | x=1.95 x=2.00 0.50 Final answer/Summation of all iterations is: 8.95 --
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